[jacquesm]

You have cause and effect switched. The field is the result of the flow of electrons, not the other way around. That makes for a nice shortcut ('it's all fields') but without that flow of electrons through a conductor carrying current there is no field. Fields are a direct result of moving charge and electrons carry charge. This is symmetrical, fields in turn have the potential to move charge. The fact that some fraction of those fields extends outside of the conductors (they have to, they are at right angles to flow of the electrons in the conductor) is not a very good reason to suddenly leap to the conclusion that all 'energy is transferred outside of the copper' until you actually use that component: such as in a transformer, or to model parasitic inductances between conductors. And then you'll find that for most wire thicknesses and most frequencies (except for special HF 'litze' wire) the bulk of the energy is still transferred within the circumference of the wire. That's because the EM field of a single electron is tiny and extends in theory out into infinity but in practice and at normal currents and frequencies is actually very limited.

The reason why people under normal circumstances would not decorate the diagram with arrows between power sources and sinks is simple: it would be redundant. Just like we don't draw capacitors, coils and resistors over every wire. We all know they are there but for most practical work you can ignore them, in fact you do your level best to ensure that those parasitic components are as small as you can make them to ensure your circuit works as intended. But once they become more dominant (which already happens at very low frequencies above DC) you have to start taking them into account, but for most regular applications you will still find that the larger part of the field is constrained within the wire and only a tiny amount extends outside of it.

So, in an extremely pedantic sense you are right: for an infinitely thin wire the EM field will lie completely outside of the wire. But for real world wires the bulk of the field is constrained within the wire.

[H8crilA]

> You have cause and effect switched. The field is the result of the flow of electrons.

Actually it's the electric field that causes the flow of electrons. The flow of electrons does not create the electric field.

It can't be the other way around, as the movement of electrons does not create an electric field, only a magnetic field (which is then crossed with the electric field outside of the conductor to create a flow of energy).

Electrons do not carry energy in a circuit. They use up some of the energy from around the conductor when they bump into atoms of the conductor, we call this process resistance. In particular they can be used purposefully to convert EM energy into heat, like in a light bulb.

PS. I keep bringing the circuit theory in because it is the reason why people can't believe how electricity actually works. But you can forget about this if it doesn't help, it's not that important.

[jacquesm]

I really hate to start quoting basic wikipedia articles but I have some other stuff to do and I don't want to leave this dangling, from:

https://en.wikipedia.org/wiki/Electric_field

This is the second sentence in that article:

"Electric fields originate from electric charges and time-varying electric currents."

> I keep bringing the circuit theory in because it is the reason why people can't believe how electricity actually works.

Circuit theory is simply a very useful approximation and how 'electricity actually works' is highly counter intuitive because for starters we got the flow of electrons wrong (they flow from - to + and not the other way around). Still, conceptually, everybody that does electronics for a living will model the flow from + to - because of convenience. But strictly speaking that's not how it 'really works'. But nobody cares they just have a job to do and the tools that suffice for that are the tools that will be used.

Finally: the QM reconciliation of EM and Gravity is for now an unsolved problem so there is a fair chance that in the long run you'll be able to make the argument that 'EM is not really how it works'. But for now it's good enough.

[colanderman]

> You have cause and effect switched. The field is the result of the flow of electrons, not the other way around.

I literally said "this field is induced by the flow of electrons." (Specifically, the magnetic field. The electric field is due to charge difference between the wires.)

> for real world wires the bulk of the field is constrained within the wire.

No, there is none - zero - nada - electric field perpendicular to the wire within said wire. If there were, the electrons would all flow to one side of the wire -- but, as you agree, they don't for DC, they are distributed throughout the bulk.

(There is a small electric field within the wire parallel to the wire; this is associated with the resistivity and energy loss in the wire, which is not related to power transfer. Its Poynting vector points outward from the wire, not to or from the source.)

[jacquesm]

So, you are proposing the the electric field of electrons moving inside a wire somehow have their fields appear outside of the wire rather than proportional to the distance from the electron (Coloumbs inverse square law)? As though the medium in which a particular electron is suspended can create an exclusion zone for its electric field?

That's interesting but contrary to just about everything that I've ever learned about this stuff. My understanding to date is that there is an electric field in a wire because if there wouldn't be then there would be a non-uniform distribution of electrons in the wire, in other words, without an electric field in the wire there would be no current to begin with, the fact that there is a current is proof that there is an electric field in the wire. I would expect that field to be aligned with the direction of the wire and constant in magnitude relative to the flow of the current. Anything else just simply does not make any sense to me.

[colanderman]

Consider the x component (along the wire's length) and the y and z components (along the wire's radius) separately.

There is a E field component along the x axis within the wire, correct. This provides the force which causes the electrons to accelerate. The field is slight -- remember it's just the voltage from one end of the wire to the other (millivolts) divided by its length. Notably, this component does not impart energy to the load -- it corresponds directly to energy dissipated in the wire itself (i.e. its Poynting vector points toward the center of the wire).

But along the wire's radius, within the wire, there's no net E field. Each electron (and each proton!) contributes a small field (which does extend immediately from the electron-point-source as you say), but the electrons arrange themselves within the metal such that the net field along these axes in the wire is zero. (If they were not arranged so, the nonzero field would push them toward this arrangement. Since the electrons aren't moving along these axes, the field along these axes is zero.) In the intuitive sense -- uniform static distribution of electrons begets a uniform static field among them.

Outside the wire -- assuming the wire has a net charge (positive or negative) -- we do see a radial electric field, because the electrons are not free to move to/from the wire through free space to neutralize this field. And if there are two wires of opposite charge (as in the circuit described), the fields sum constructively (as one wire's field is radially outward, the other's radially inward). This field is much larger, particularly if the wires are physically near each other, since it derives from the voltage difference between the wires, rather than across them. (And most importantly, the Poynting vector of this field points toward the load.)

You may be right that the region outside the wire where energy flows may be small and near the wire (working through some math now) but it's definitely outside the wire for the reasons given above, and flows between sources and sinks according to the Poynting field, not along any path that's visually represented in a circuit diagram.

[jacquesm]

> There is a E field component along the x axis within the wire, correct. This provides the force which causes the electrons to accelerate. The field is slight -- remember it's just the voltage from one end of the wire to the other (millivolts) divided by its length. Notably, this component does not impart energy to the load -- it corresponds directly to energy dissipated in the wire itself (i.e. its Poynting vector points toward the center of the wire).

Ok so far that is in accordance with how I understand things to work, assuming a 'good conductor' and low impedance for AC or varying current.

> But along the wire's radius, within the wire, there's no net E field.

Because there are no electrons flowing in that direction, and even if there were there would be others canceling that out (this holds on average for thermal noise as well, which would show up as minor current fluctuations), to get an E field at right angles to the wire you'd have to to see a voltage difference between the center of the wire and the outside of the wire, and barring local effects there will be no such thing. Essentially because there is a dead short between those two points (a very short piece of a section of the wire) Right?

> Each electron (and each proton!) contributes a small field (which does extend immediately from the electron-point-source as you say), but the electrons arrange themselves within the metal such that the net field along these axes in the wire is zero. (If they were not arranged so, the nonzero field would push them toward this arrangement. Since the electrons aren't moving along these axes, the field along these axes is zero.) In the intuitive sense -- uniform static distribution of electrons begets a uniform static field among them.

Except for the limit case of a single electron moving in a conductor, that presumably would result in a net (extremely small) field, but for larger numbers of electrons (any practical measurable current) that would all cancel out within the wire perimeter.

> Outside the wire -- assuming the wire has a net charge (positive or negative) -- we do see a radial electric field, because the electrons are not free to move to/from the wire through free space to neutralize this field.

Because the wire perimeter is defined as the area where the conductor ends (which makes me wonder how that transition would work in a material that consists of an outer core that gradually becomes higher impedance).

> This field is much larger,

Right, because it can't neutralize because it is in a dielectricum.

> particularly if the wires are physically near each other

Yes, but that's not a given and any old wire loop has its wires as far away from each other as you want them to be. Coaxial cable is the way it is to get the fields to cancel out so that the amount of spurious radiation from the cable is kept to a minimum (you can also twist the two wires around each other for some of the same effect).

> since it derives from the voltage difference between the wires, rather than across them.

You mean 'wires on both sides of the load'? Somewhere along the line we switched from discussing a single wire in isolation to two wires.

> (And most importantly, the Poynting vector of this field points toward the load.)

So we're now talking about the electrical field between the two wires on either side of the load?

> You may be right that the region outside the wire where energy flows may be small and near the wire

I don't see how it could be any other way but I'm curious as to the results of your calculations.

> but it's definitely outside the wire for the reasons given above

Ok

> and flows between sources and sinks according to the Poynting field, not along any path that's visually represented in a circuit diagram.

Ok, I think I'm finally seeing what you are getting at, you are talking about the net flow of energy from a 'source' (say a battery or a generator) to a 'load' and that energy clearly flows in one direction even if the electrons themselves make a round trip from one side of the source to the other side of the source (well, technically they don't even do that, they may well reverse direction before they get there), and that the field that carries that energy is the field at right angles to the wire rather than the field that you see across the length of the wire.

edit: thinking about this some more: essentially the two wires you are talking about have some of the properties of a charged capacitor, and the electric field is passing between those two 'plates'. That makes me wonder if the field is mostly concentrated between the 'plane' (it can be quite a complex shape) between the two conductors rather than that most of it is in free space radiating outward. After all an electric field normally does the same thing for a charged capacitor, the bulk of the field sits between the plates with only a small fraction moving outside of the envelope of the plates.

[jacquesm]

Yet another edit, much more thinking later: isn't it the electromagnetic field then (that consists of photons) that moves the energy from source to sink because photons are capable of moving energy rather than the electric field between the wires? After all you'd need something moving energy and electromagnetism can transfer energy whereas an electric field (or even a moving charge) does not? (as in: current does not equal 'work', that all depends on the associated voltage). It also would explain the 'single electron' issue outlined above, that wouldn't apply, you get a magnetic fieldline or you don't and if you do that single photon can do useful work.

[colanderman]

I think we're mostly on the same page, yes, sorry I wasn't clear, I've been talking about two wires, linking a power source with a load.

> isn't it the electromagnetic field then (that consists of photons) that moves the energy from source to sink because photons are capable of moving energy rather than the electric field between the wires?

Yes, that's my understanding also; I focused on the electric field in my last post because that's what differs inside and outside a conductor; the magnetic field exists and has the same orientation both in and outside a conductor carrying current. But it's the product of the two which defines the Poynting vector, which indicates the flow of energy in the EM field.

What I've calculated so far (for the case of two infinite parallel wires) does indicate that the Poynting vector is concentrated around the wires -- it drops off approximately with the 4th power of the distance from either wire (though there's a decent saddle between them). I plan still to calculate some integrals around the wires because I'm curious actual numbers. For reference the expression I've derived is:

|S| = Pr² / (ln(2r/a) × ((y² - z² - r²)² + 4y²z²))

where:

S = Poynting vector

P = power delivered

2r = distance between wires

a = diameter of wire (this only creeps in when relating voltage between the wires to charge on the wires)

The Poynting field points uniformly from the source to the sink, in the +x direction (same axis as the wires). The y axis points from one wire to the other.